Applying the induction hypothesis indirectly? I have trouble understanding this proof.












3












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I am confused about the mechanics of the following proof (page 423, chapter 20, of the fourth edition of Spivak's Calculus (Taylor's Theorem)):



enter image description here



I am not sure that I understand how applying the induction hypothesis to $R'$ works in proving the lemma for $R$ for all $n$. I expected that the induction hypothesis would be applied to $R$.



Maybe my confusion stems from the fact that I don't apply the induction hypothesis generally enough. Maybe I need to assume that it holds for some $n$ for all functions $f$ that meet the given criteria and that it therefore also holds for $f'$, from which the case $n+1$ follows for all $f$. So we assume that the induction hypothesis is true for both $R$ and $R'$ but use the latter to derive the $n+1$ case for the former. This also requires that $R'$ is $(n)$-times differentiable with $R'^{(k)}(a) = 0$ for all k up to $n-1$. But this appears always to be the case.






real-analysis calculus induction taylor-expansion







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    3












    $begingroup$


    I am confused about the mechanics of the following proof (page 423, chapter 20, of the fourth edition of Spivak's Calculus (Taylor's Theorem)):



    enter image description here



    I am not sure that I understand how applying the induction hypothesis to $R'$ works in proving the lemma for $R$ for all $n$. I expected that the induction hypothesis would be applied to $R$.



    Maybe my confusion stems from the fact that I don't apply the induction hypothesis generally enough. Maybe I need to assume that it holds for some $n$ for all functions $f$ that meet the given criteria and that it therefore also holds for $f'$, from which the case $n+1$ follows for all $f$. So we assume that the induction hypothesis is true for both $R$ and $R'$ but use the latter to derive the $n+1$ case for the former. This also requires that $R'$ is $(n)$-times differentiable with $R'^{(k)}(a) = 0$ for all k up to $n-1$. But this appears always to be the case.






    real-analysis calculus induction taylor-expansion







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I am confused about the mechanics of the following proof (page 423, chapter 20, of the fourth edition of Spivak's Calculus (Taylor's Theorem)):



      enter image description here



      I am not sure that I understand how applying the induction hypothesis to $R'$ works in proving the lemma for $R$ for all $n$. I expected that the induction hypothesis would be applied to $R$.



      Maybe my confusion stems from the fact that I don't apply the induction hypothesis generally enough. Maybe I need to assume that it holds for some $n$ for all functions $f$ that meet the given criteria and that it therefore also holds for $f'$, from which the case $n+1$ follows for all $f$. So we assume that the induction hypothesis is true for both $R$ and $R'$ but use the latter to derive the $n+1$ case for the former. This also requires that $R'$ is $(n)$-times differentiable with $R'^{(k)}(a) = 0$ for all k up to $n-1$. But this appears always to be the case.






      real-analysis calculus induction taylor-expansion







      share|cite|improve this question









      $endgroup$




      I am confused about the mechanics of the following proof (page 423, chapter 20, of the fourth edition of Spivak's Calculus (Taylor's Theorem)):



      enter image description here



      I am not sure that I understand how applying the induction hypothesis to $R'$ works in proving the lemma for $R$ for all $n$. I expected that the induction hypothesis would be applied to $R$.



      Maybe my confusion stems from the fact that I don't apply the induction hypothesis generally enough. Maybe I need to assume that it holds for some $n$ for all functions $f$ that meet the given criteria and that it therefore also holds for $f'$, from which the case $n+1$ follows for all $f$. So we assume that the induction hypothesis is true for both $R$ and $R'$ but use the latter to derive the $n+1$ case for the former. This also requires that $R'$ is $(n)$-times differentiable with $R'^{(k)}(a) = 0$ for all k up to $n-1$. But this appears always to be the case.






      real-analysis calculus induction taylor-expansion




      real-analysis calculus induction taylor-expansion






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      asked 7 hours ago









      user245312user245312

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          $begingroup$

          Yes, the statement $P(n)$ that you prove is




          For every $n+1$ times differentiable function $R$, ...




          and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$.
          For induction, you show $P(0)$, i.e.,




          For every (once) differentiable function $R$, ...




          and you show $forall ncolon P(n)to P(n+1)$ to finally arrive at




          For all $ninBbb N_0$, for every $n+1$ times differentiable function $R$ ...




          For the induction step $forall ncolon P(n)to P(n+1)$, we proceed as follows:
          Let $ninBbb N_0$ be arbitrary. We want to show $P(n)to P(n+1)$.
          So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)to P(n+1)$. As $n$ was arbitrary, we conclude $forall ninBbb N_0colon P(n)to P(n+1)$.






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            $begingroup$

            We have taken the function $R $ such that it is $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$,
            $k=0,1,ldots,n+1$, we have $(R')^{(k)}(a)=0$,
            $k=0,1,ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook.






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              2 Answers
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              $begingroup$

              Yes, the statement $P(n)$ that you prove is




              For every $n+1$ times differentiable function $R$, ...




              and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$.
              For induction, you show $P(0)$, i.e.,




              For every (once) differentiable function $R$, ...




              and you show $forall ncolon P(n)to P(n+1)$ to finally arrive at




              For all $ninBbb N_0$, for every $n+1$ times differentiable function $R$ ...




              For the induction step $forall ncolon P(n)to P(n+1)$, we proceed as follows:
              Let $ninBbb N_0$ be arbitrary. We want to show $P(n)to P(n+1)$.
              So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)to P(n+1)$. As $n$ was arbitrary, we conclude $forall ninBbb N_0colon P(n)to P(n+1)$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Yes, the statement $P(n)$ that you prove is




                For every $n+1$ times differentiable function $R$, ...




                and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$.
                For induction, you show $P(0)$, i.e.,




                For every (once) differentiable function $R$, ...




                and you show $forall ncolon P(n)to P(n+1)$ to finally arrive at




                For all $ninBbb N_0$, for every $n+1$ times differentiable function $R$ ...




                For the induction step $forall ncolon P(n)to P(n+1)$, we proceed as follows:
                Let $ninBbb N_0$ be arbitrary. We want to show $P(n)to P(n+1)$.
                So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)to P(n+1)$. As $n$ was arbitrary, we conclude $forall ninBbb N_0colon P(n)to P(n+1)$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Yes, the statement $P(n)$ that you prove is




                  For every $n+1$ times differentiable function $R$, ...




                  and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$.
                  For induction, you show $P(0)$, i.e.,




                  For every (once) differentiable function $R$, ...




                  and you show $forall ncolon P(n)to P(n+1)$ to finally arrive at




                  For all $ninBbb N_0$, for every $n+1$ times differentiable function $R$ ...




                  For the induction step $forall ncolon P(n)to P(n+1)$, we proceed as follows:
                  Let $ninBbb N_0$ be arbitrary. We want to show $P(n)to P(n+1)$.
                  So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)to P(n+1)$. As $n$ was arbitrary, we conclude $forall ninBbb N_0colon P(n)to P(n+1)$.






                  share|cite|improve this answer









                  $endgroup$



                  Yes, the statement $P(n)$ that you prove is




                  For every $n+1$ times differentiable function $R$, ...




                  and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$.
                  For induction, you show $P(0)$, i.e.,




                  For every (once) differentiable function $R$, ...




                  and you show $forall ncolon P(n)to P(n+1)$ to finally arrive at




                  For all $ninBbb N_0$, for every $n+1$ times differentiable function $R$ ...




                  For the induction step $forall ncolon P(n)to P(n+1)$, we proceed as follows:
                  Let $ninBbb N_0$ be arbitrary. We want to show $P(n)to P(n+1)$.
                  So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)to P(n+1)$. As $n$ was arbitrary, we conclude $forall ninBbb N_0colon P(n)to P(n+1)$.







                  share|cite|improve this answer












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                  answered 7 hours ago









                  Hagen von EitzenHagen von Eitzen

                  279k23271502




                  279k23271502























                      3












                      $begingroup$

                      We have taken the function $R $ such that it is $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$,
                      $k=0,1,ldots,n+1$, we have $(R')^{(k)}(a)=0$,
                      $k=0,1,ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook.






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        We have taken the function $R $ such that it is $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$,
                        $k=0,1,ldots,n+1$, we have $(R')^{(k)}(a)=0$,
                        $k=0,1,ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook.






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          We have taken the function $R $ such that it is $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$,
                          $k=0,1,ldots,n+1$, we have $(R')^{(k)}(a)=0$,
                          $k=0,1,ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook.






                          share|cite|improve this answer









                          $endgroup$



                          We have taken the function $R $ such that it is $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$,
                          $k=0,1,ldots,n+1$, we have $(R')^{(k)}(a)=0$,
                          $k=0,1,ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 7 hours ago









                          Thomas ShelbyThomas Shelby

                          3,1071523




                          3,1071523






























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